Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int K = 19;
vector<int> v[N], p[N][K];
int h[N], anc[N][K];
vector<int> merge(vector<int> a, vector<int> b) {
  if (!a.size()) return b;
  if (!b.size()) return a;
  vector<int> ret(a.size() + b.size());
  merge(a.begin(), a.end(), b.begin(), b.end(), ret.begin());
  vector<int>::iterator it = unique(ret.begin(), ret.end());
  ret.resize(min((int)distance(ret.begin(), it), 10));
  return ret;
}
void dfs(int s, int P) {
  anc[s][0] = P;
  if (P + 1) h[s] = h[P] + 1;
  for (int i = 1; i < K; i++)
    if (anc[s][i - 1] + 1) {
      anc[s][i] = anc[anc[s][i - 1]][i - 1];
      p[s][i] = merge(p[s][i - 1], p[anc[s][i - 1]][i - 1]);
    }
  for (auto i : v[s])
    if (i - P) dfs(i, s);
}
vector<int> LCA(int x, int y) {
  vector<int> ret;
  if (h[x] > h[y]) swap(x, y);
  for (int i = K - 1; i >= 0; i--)
    if (anc[y][i] + 1 && h[anc[y][i]] >= h[x]) {
      ret = merge(ret, p[y][i]);
      y = anc[y][i];
    }
  if (x == y) {
    ret = merge(ret, p[y][0]);
    return ret;
  }
  for (int i = K - 1; i >= 0; i--)
    if (anc[x][i] - anc[y][i]) {
      ret = merge(ret, merge(p[x][i], p[y][i]));
      x = anc[x][i], y = anc[y][i];
    }
  return (merge(ret, merge(p[x][1], p[y][1])));
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 1; i <= m; i++) {
    int x;
    cin >> x;
    p[x][0].push_back(i);
  }
  for (int i = 0; i < N; i++) sort(p[i][0].begin(), p[i][0].end());
  memset(anc, -1, sizeof(anc));
  dfs(1, -1);
  while (q--) {
    int x, y, a;
    cin >> x >> y >> a;
    vector<int> ret = LCA(x, y);
    int yo = min((int)ret.size(), a);
    cout << yo << " ";
    for (int i = 0; i < yo; i++) cout << ret[i] << " ";
    cout << "\n";
  }
  return 0;
}